1. Let 0 ≠ a = bc. If a b, say b = ua, then a = uac so 1 = uc, c is a unit, c 1. If c 1 then c is a unit so a = bc b.

3. a. 2 + i = i(1 − 2i) in and i is a unit.

5. Let a a′, b b′, say a′ = ua, b′ = vb. Then a′b′ = (uv)(ab) so a′b′ ab.

7. If is a unit, 1 = N(u)N(u⁻¹) = (a² + 5b²) · N(u⁻¹) shows a² + 5b² = 1. Thus u = ± 1.

9. If is maximal in P = { a ∉ R, a ∉ R^∗}, let 83 p = ab. Then p ⊆ a so either a = p or a ∉ R^∗; that is a p or a 1. Thus p is irreducible. Conversely, if p is irreducible and p ⊆ a, a ∉ R^∗, then p = ab where b is a unit. Hence a = pb⁻¹ p , so p = a.

10.

11. If p ≡ 3 (mod 4), suppose p = ab in . Then p² = |a|²|b|². Since |x| = 1 in means x is a unit, it suffices to show that |a|² = p is impossible in if a = m + in. This means m² + n² = p, whence m² + n² = 0 in . If (say) m ≠ 0 in , this gives 1 + (m⁻¹n)² = 0 in . This cannot happen (Corollary to Theorem 8 §1.3) because p ≡ 3 (mod 4). Hence m = 0 and a = 0 in ; that is p|m and p|n. But then m² + n² = p implies p|1, a contradiction.

12.

13. a. If , suppose p = ab in . Then 9 = N(p) = N(a)N(b). But N(a) = 3 is impossible in (because m² + 5n² ≠ 3 for m, , so N(a) = 1 or N(b) = 1. Thus a = ± 1 or b = ± 1, and we have shown that p is irreducible in . To see that it is not prime, the fact that N(p) = 9 gives . So if p is prime, then p|3 in , say 3 = p · q. Then 9 = N(3) = N(p)N(q) = 9N(q) so N(q) = 1. This means q = ± 1, whence , a contradiction. So p is not prime.

14.

15. If then p = ab means 4 = N(p) = N(a)N(b). Since m² + 3n² = 2 is impossible for m, , we have N(a) ≠ 2 and N(b) ≠ 2. So N(a) = 1 or N(b) = 1; that is a = ± 1 or b = ± 1. Thus p is irreducible. If p is prime then shows that p|2 in , say 2 = pq. Thus 4 = N(p)N(q) = 4N(q), whence N(q) = 1, q = ± 1, 2 = ± p, a contradiction. So p is not prime.

16. a. If p q, suppose p is irreducible. If q = ab in R then p ab so p a or p b. Thus q a or q b, so q is irreducible. The converse is proved the same way.

17. Assume and N(p) is as in Example 3. If N(p) is a prime and p = xy in R, then N(p) = N(x)N(y). It follows that N(x) = 1 or N(y) = 1, say N(x) = 1. If , this means m² + 5n² = 1, whence x = ± 1 is a unit. This shows that p is irreducible.

19. Suppose R is an integral domain with the DCCP. If a ≠ 0 in R, we have a ⊇ a² ⊇ a³ so, by hypothesis, there exists a ≥ 1 such that aⁿ = aⁿ⁺¹. Thus aⁿ aⁿ⁺¹ = Raⁿ⁺¹, say aⁿ = baⁿ⁺¹. Since a ≠ 0 and R is a domain, this gives 1 = ba, so a is a unit. This shows that R is a field. The converse is clear as a field has only two ideals.

21. If R has the ACCP, suppose F has no maximal member. Choose a₁ inF. Then a₁ is not maximal so a₁ ⊂ a₂ for some a₂ in F. Now a₂ is not maximal, so a₂ ⊂ a₃ for some a₃ in F. This continues to violate the ACCP. Conversely, suppose a₁ ⊆ a₂ ⊆ and let F = { a_i i = 1, 2, . . . }. By hypothesis, let a_n be maximal in F. If m ≥ n then a_n ⊆ a_m so the maximality gives a_n = a_m. Thus R has the ACCP by Lemma 1.

23. No. is a subring of the UFD which is itself not a UFD by Example 5 (and Theorem 7).

24. a Write d = gcd (a, b, . . .). Then d|0, d|a, d|b, . . .; if r|0, r|a, r|b, . . . then r|d by definition. So d gcd (0, a, b, . . .) . Clearly 0|0, a|0, b|0, . . .; if 0|r, a|r, b|r, . . . then r = 0 (because 0|r) so 0 lcm(0, a, b, . . .) by definition.

25. a Write d = gcd (a₁, . . ., a_n) and d′ = gcd (b₁, . . ., b_n). Then d|a_i for all i so d|b_i for all i (Exercise 2), whence d|d′. Similarly, d′|d, whence d′ d.

27. Write d = gcd [a, gcd (b, c)] and d′ = gcd [gcd (a, b), c]. Then d divides a and gcd (b, c) so it divides all of a, b, c. Thus d divides gcd (a, b) and c, whence d|d′. Similarly d′|d so d d′. Moreover we have shown that d|a, d|b, d|c. If r|a, r|b and r|c then r divides a and gcd (b, c) so r|d. Hence gcd (a, b, c, ) exists and d gcd (a, b, c).

29. It is clear if c = 0. Let , and . Then a|bc means a_i ≤ b_i + c_i for all i, and gcd (a, b) 1 means min (a_i, b_i) = 0 for all i. For each i, min (a_i, b_i) = 0 means a_i = 0 or b_i = 0. Thus a_i ≤ c_i in either case (because a_i ≤ b_i + c_i) so a|c.

31. If m lcm(a₁, . . ., a_n) exists in R then a_i|m for each i shows m ⊆ a_i for each i, and hence that m ⊆ A where we write A = a_i ∩ ∩ a_n. But r A means a_i|r for each i, so m|r by definition. Thus r m and we have m = A. Conversely, if A = m then a_i|m for each i (because m a_i); and, if a_i|r for each i, then r A = m so m|r. Thus m is a least common multiple of the a_i.

33. 1.If the nonzero coefficients of f are a₁, a₂, . . ., a_n, then

34.

35. Suppose f = gh, h R[x]. Then Gauss' lemma gives c(g)c(h) c(f) 1. Hence c(g)c(h) is a unit, so c(g) is a unit. This means g is primitive.

37. If f g in R[x] then f = ug, u a unit in R[x]. Thus u is a unit in F[x] so f g in F[x]. Conversely, let f = ug, u a unit in F[x]. Then u F, say . Thus bf = ag in R so, since f and g are primitive,

39. a. We have . This is a subring of because

40.